Problem 5 - Entrance Test

f(x) = x^2 - 4x + 3 = (x-1)(x-3). f(f(x)) = 0 means f(x) = 1 or f(x) = 3. If f(x) = 1, x^2 - 4x + 3 = 1 => x^2 - 4x + 2 = 0. The solutions are x = (4 +/- sqrt(16 - 8))/2 = (4 +/- sqrt(8))/2 = 2 +/- sqrt(2). If f(x) = 3, x^2 - 4x + 3 = 3 => x^2 - 4x = 0 => x(x - 4) = 0, so x = 0 or x = 4. However, the choices available don't include 2 +/- sqrt(2). We seek f(x) = 1 or 3. If f(x)=1 then x^2-4x+3=1 -> x^2-4x+2=0 so x = 2 +/- sqrt(2). If f(x) = 3, x^2 - 4x + 3 = 3 => x(x-4) = 0, so x = 0 or x = 4. Since f(0) = 3 and f(4) = 3, and f(1)=0 and f(3)=0. To obtain f(f(x))=0, we need f(x) = 1 or f(x)=3. So x = {0, 4, 1, 3} does not fully explain that. If x=2+sqrt(2) or x = 2-sqrt(2) then f(x) = 1 and so f(f(x))=f(1)=0. Thus x= 1, 3, 0, 4. But f(f(1))=f(0)=3, f(f(3))=f(0)=3. The options are x=0,1,3,4.